OD-Characterization of Alternating and Symmetric Groups of Degree p+5?

Yanxiong YANHaijing XUGuiyun CHEN

1 Introduction

Throughout this paper,groups under consideration are finite and simple groups that are non-Abelian.For any groupG,we useπe(G)to denote the set of orders of its elements and denote byπ(G)the set of prime divisors of|G|.We associate toπ(G)a graph ofG,denoted by Γ(G)(see[1]).The vertex set of this graph isπ(G),and two distinct verticesp,qare adjacent by an edge if and only ifpq∈πe(G).In this case,we writep～q.We also denote byπ(n)the set of all primes dividingn,wherenis a positive integer.

In this article,we use the following symbols.For a finite groupG,the socle ofGis de fined as the subgroup generated by minimal normal subgroups ofG,denoted as Soc(G).Sylp(G)denotes the set of all Sylowp-subgroups ofG,wherep∈π(G),andPrdenotes a Sylowrsubgroup ofGforr∈π(G).Furthermore,the symmetric and alternating groups of degreenare denoted bySnandAn,respectively.Letqbe a prime,and we use Exp(m,q)to denote the exponent of the largest power of a primeqin the factorization of a positive integerm(>1).All further unexplained symbols and notations are standard and can be found in[2],for instance.

Definition 1.1(see[3])Let G be a finite group and|G|=where pis areprimes and αis are integers.For p∈π(G),letdeg(p):=|{q∈π(G)|p～q}|,which we call the degree of p.We also de fine D(G):=(deg(p1),deg(p2),...,deg(pk)),where p1

Definition 1.2(see[3])A group M is called k-fold OD-characterizable if there exist exactly k non-isomorphic groups G such that(1)|G|=|M|and(2)D(G)=D(M).Moreover,a1-fold OD-characterizable group is simply called an OD-characterizable group.

It is an interesting and difficult topic to determine the structure of finite groups by their orders and degree patterns.This topic is related to the following open problem.

Open problem(see[3]) LetGandMbe finite groups satisfying the conditions(1)|G|=|M|and(2)D(G)=D(M).Then

(i)How far do these conditions af f ect the structure ofG?

(ii)Does the number of non-isomorphic groups satisfy(1)and(2)finite?

At present,we mention that the problem is still unsolved completely and till now we may not be able to provide a suitable answer for the above questions.This topic was studied in several articles.For example,in a series of articles(see[3–20]),it was shown that many finite almostsimple groups arem-foldOD-characterizable,wheremis a positive integer andm≥1.For convenience,we summarize some results of these articles in the following Propositions 1.1–1.4.

Proposition 1.1(see[3–12])Let p be a prime.A finite group G is OD-characterizable if G is isomorphic to one of the following groups:

(1)The alternating groups Ap,Ap+1and Ap+2;

(2)the alternating groups Ap+3,where7p∈π(100!);

(3)all finite almost simple K3-groups exceptAut(A6)andAut(U4(2));

(4)the symmetric groups Spand Sp+1;

(5)all finite simple K4-groups except A10;

(6)all finite simple C2,2-groups;

(7)the simple groups of Lie type L2(q),L3(q),U3(q),2B2(q)and2G2(q)for certain prime power q;

(8)all sporadic simple groups and their automorphism groups exceptAut(J2)andAut(McL);

(9)the almost simple groups:Aut(F4(2)),Aut((2))andAut();

(10)almost simple group L2(49)·2i(i=1,2,3);

(11)projective general linear group PGL(2,q)for certain odd prime power q;

(12)all finite simple groups whose orders are less than108except for A10and U4(2).

Proposition 1.2(see[13–16])A finite group G is2-fold OD-characterizable if G is one of the following groups:A10,U4(2),S6(3),O7(3),B2(q),Cp(3),almost simple groups2·F4(2),Aut(J2)andAut(McL).

Proposition 1.3(see[17–20])A finite group G is3-fold OD-characterizable if G is one of the following groups:

(1)The almost simple groups of U3(5)·3and U6(2)·3;

(2)the symmetric groups Sn,where n≤100and n10,p,p+1.

Proposition 1.4(see[13])Let G be a finite group with|G|=|S10|and D(G)=D(S10),then G is8-fold OD-characterizable.

2 Main Results

According to Proposition 1.1(1),the alternating groupsAp,Ap+1andAp+2areOD-characterizable,andAp+3with 7p∈π(100!)isOD-characterizable.Proposition 1.2 says that the alternating groupA10is 2-foldOD-characterizable.Now,omitting all the above alternating groups exceptA10,there remain the following alternating groups:

By[1],it is easy to check that all the groups in(2.1)have the connected prime graph.By Proposition 1.1,we see that it is difficult to investigate how many-foldOD-characterization of alternating groups are.In this paper,we continue to investigate this topic and get the following theorem.

Theorem 2.1All alternating groups Ap+5,where p+4is a composite number and p+6is a prime andp∈π(1000!),are OD-characterizable.

Proposition 1.2 says thatA10is 2-foldOD-characterizable.It is worth mentioning thatA10is the first alternating group which has not been considered asOD-characterizable.Up to now,we have not found an alternating groupAn(n10)which is notOD-characterizable.Hence,we put forward the following question.

Question 2.1Are the alternating groupsAn(n10)OD-characterizable?

In fact,Theorem 2.1 and Proposition 1.1(1)–(3)imply the following corollary.

Corollary 2.1Let Anbe an alternating group of degree n.Assume that one of the following conditions is fulfilled:

(1)n=p,p+1or p+2,where p is a prime;

(2)n=p+3,where7p∈π(100!);

(3)n=p+5,where p+4is a composite number and p+6is a prime and5p∈π(1000!).Then Anis OD-characterizable.

In this article,we will also show the following theorem.

Theorem 2.2All symmetric groups Sp+5,where p+4is a composite number and p+6is a prime and5p∈π(1000!),are3-fold OD-characterizable.

3 Preliminaries

In this section,we give some results which will be applied for our further investigations.We shall utilize the following Lemma 3.1 concerning the set of elements of the alternating and symmetric groups(see[21]).

Lemma 3.1(see[21])The group Sn(or An)has an element of order m=where p1,p2,···,psare distinct primes and α1,α2,···αsare natural numbers,if and only if≤n for m odd,andfor m even).

As an immediate corollary of Lemma 3.1,we have the following lemma.

Lemma 3.2Let An(or Sn)be an alternating group(or a symmetric group)of degree n.Then the following assertions hold:

(1)Let p,q∈π(An)be odd primes.Then p～q if and only if p+q≤n.

(2)Let p∈π(An)be an odd prime.Then2～p if and only if p+4≤n.

(3)Let p,q∈π(Sn).Then p～q if and only if p+q≤n.

Lemma 3.3(see[22])Let G be a finite solvable group,all of whose elements are of prime power order.Then|π(G)|≤ 2.

Lemma 3.4Let Ap+5be an alternating group of degree p+5,where p is a prime,and p+2and p+4are composite numbers.Suppose that|π(Ap+5)|=d.Then the following assertions hold:

(1)deg(2)=d?1.Particularly,2～r for each r∈π(Ap+5).

(2)deg(3)=deg(5)=d?1,i.e.,3～r and5～r for each r∈π(Ap+5).

(3)deg(p)=3.In other words,p～r,where r∈ π(Ap+5),if and only if r=2,r=3or r=5.

(4)Exp(|Ap+5|,2)=In particular,Exp(|Ap+5|,2)

(5)Exp(|Ap+5|,r)=for each r∈π(A){2}.Furthermore,Exp(|A|,r)thenExp(|Ap+5|,r)=1.

ProofBy Lemma 3.2,one has that 2p∈ πe(Ap+5).Clearly,sincer+4≤p+5 for anyr∈π(Ap+5){p},it follows that 2～r,so deg(2)=d?1.Similarly,we have deg(3)=deg(5)=d?1.Forr∈π(Ap+5){2,p},by Lemma 3.2,it is easy to check thatp～rif and only ifr≤5.Hencer=2,3 or 5.Thus deg(p)=3.Till now we have proved that(1)–(3)hold.

By the definition of Gauss’s integer function,we have that

Hence Exp(|Ap+5|,2)

Again,using Gauss’s integer function,we have that Exp(|Ap+5|,r)=Therefore Exp(|Ap+5|,r)then Exp(|Ap+5|,r)=1.Hence(5)follows.This completes the proof of Lemma 3.4.

Lemma 3.5(see[23])Let a be an arbitrary integer and m a positive integer.If(a,m)=1,then the equation ax≡1(modm)has solutions.Moreover,if the order of a modulo m is h(a),then h(a)|?(m),where ?(m)is an Euler’s function of m.

Lemma 3.6Let Ap+5be an alternating group of degree p+5,where p+4is a composite number,p+6is a prime and100

(i)where s(q)=Exp(|Ap+5|,q).

(ii)If p∈{131,167,173,233,251,257,263,373,383,433,503,541,557,563,587,607,647,677,727,733,941,977},then p|NG(Q)|.

(iii)If p∈{157,271,331,353,367,443,571,593,601,653,751,947,971},then there exists at least a prime number,say r,such that the order of r modulo p is less than p?1,where7≤r

ProofObviously,the equationqx≡1(modp)has solutions by Lemma 3.5.Suppose that the order ofqmodulopish(q).Ifh(q)=p?1,then we callqa primitive root of modulop.By hypothesis,it is easy to check that there are only 35 such groups satisfying the conditions thatp+4 is a composite number,p+6 is a prime number and 100

Table 1 (p and h(q))

Now,using the N-C theorem,the factor groupNG(P)/CG(P)Aut(P)Hence,|NG(P)/CG(P)||(p?1).By Lemma 3.4(3),one has thatπ(NG(P))?{2,3,5}∪π(p?1).By Table 1,if there exists a prime,sayq,where 7≤q

Assume thatp∈{131,167,173,233,251,257,263,373,383,433,503,541,557,563,587,607,647,677,727,733,941,977}.Ifp||NG(Q)|,by Table 1 and Exp(|Ap+5|,q)<

Lemma 3.7(see[2,24])Let M be a finite non-Abelian simple group with order having prime divisors at most997.Then M is isomorphic to one of the simple groups listed in Tables1–3in[24].Particularly,if|π(Out(M))1,then π(Out(M))?{2,3,5,7}.

Lemma 3.8(see[25])Let S=P1×P2×···×Pr,where Pi(i=1,2,···,r)are isomorphic non-Abelian simple groups.ThenAut(S)=(Aut(P1)×Aut(P2)×···×Aut(Pr))Sr,where Sris a symmetric group of degree r.

4 OD-Characterization of the Alternating Group Ap+5

In this section,we will prove the following Theorem 2.1.It is worth mentioning that this result not only generalizes the results in[4]but also gives an affirmative answer to the Question 1.1 of this article for the alternating groupAp+5.

Proof of Theorem 2.1LetGbe a finite group satisfying the conditions that(1)|G|=|Ap+5|and(2)D(G)=D(Ap+5),wherep+4 is a composite number,p+6 is a prime andp∈π(1000!).By[17],we only need to discuss the alternating groupsAp+5,wherep+4 is a composite number,p+6 is a prime and 100p+5}∩πe(G)=?,where 2r,s∈π(G).By Lemma 3.4,the prime graph ofGis connected since deg(2)=|π(G)|? 1.Moreover,by the structure of the degree patternD(G),it is easy to check by Magma of computation group software that Γ(G)= Γ(Ap+5).In the following,we shall prove thatFor convenience,we divide the proof of Theorem 2.1 into three separate cases.

Case 1LetKbe the maximal normal solvable subgroup ofG.ThenKis a{2,3,5}-group.In particular,Gis a nonsolvable group.

We first show thatKis a-group.Assume to the contrary,and letpdivide the order ofK.SetP∈Sylp(G).By Lemma 3.6(i),we haveqs(q)|NG(P)|for anyq∈π(G)and 7≤qp,so thenrq.SinceKis solvable,it possesses a Hall{p,q,r}-subgroupT.It follows thatTis solvable.Since there exists no edge between any two distinct vertices ofp,qandrin Γ(G),all elements inTare of prime power order.Hence|π(T)|≤ 2 by Lemma 3.3,a contradiction.HenceKis a-group.

We shall argue next thatKis a-group for anyq∈ π(G){2,3,5,p}.SetQ∈Sylq(K),whereq∈π(K).Suppose that the order ofqmodulopish(q).By the Frattini argument,G=KNG(Q),and hencepdivides the order ofNG(Q).By Lemma 3.6(ii)–(iii),it follows thatpcan only be equal to one of the primes:157,271,331,353,367,443,571,593,601,653,751,947 and 971.In this case,there necessarily exists at least a prime,sayq,such thath(q)

Subcase 1.1To prove the case follows ifp=157.

By Table 1,If there exists a primeqsuch thatp||NG(Q)|,whereQ∈Sylq(G),thenq=13.By the N-C theorem,Aut(Q).By Lemma 3.4(5),we have Exp(|G|,13)=12,and thus|NG(Q)/CG(Q)||1366·(13i?1).By Magma,it is easy to check that 1511366·(13i?1).If 151||NG(Q)|,then 151∈π(CG(Q)).Thus 13～ 151,a contradiction.Hence 151∈π(K).SinceKis solvable,it possesses a Hall{13,151}-subgroupHof order 1312·151.Clearly,His nilpotent,so 13～ 151,a contradiction.

Subcase 1.2To prove the case follows ifp=271.

In this case,we know that there exists a prime,sayq,such thatp||NG(Q)|,whereQ∈Sylq(G).Thenq=29 by Table 1.On the other hand,the factor groupNG(Q)/CG(Q)is isomorphic to a subgroup of Aut(Q)by N-C theorem and Exp(|G|,29)=9 by Lemma 3.4,so|NG(Q)/CG(Q)||2936·(29i?1).It is easy to check that 2692936·(29i?1).If 269||NG(Q)|,then 269∈ π(CG(Q)).Thus 269～29,a contradiction.Hence 269∈ π(K).SinceKis solvable,it possesses a Hall{29,269}-subgroupHof order 299·269.Clearly,His Abelian,so 29～269,a contradiction.

Subcase 1.3To prove the case follows ifp=331.

In the case,there exists a prime,sayq,such thatp||NG(Q)|,whereQ∈Sylq(G).Thenq=31 by Table 1.On the other hand,we have thatNG(Q)/CG(Q)Aut(Q)by the N-C theorem and Exp(|G|,31)=10 by Lemma 3.4,so|NG(Q)/CG(Q)||3145·(31i?1).It is easy to compute thatIf 47||NG(Q)|,then 47∈π(CG(Q)).SetN=NG(Q),C=CG(Q)andK47∈Syl47(CG(Q)).By Lemma 3.4,we have Exp(|G|,47)=7.Again,by the Frattini argument,we have thatN=CNN(K47)and henceNN(K47)|.Thusp||CG(Q)|,so deg(p)≥4,a contradiction.Therefore 47|NG(Q)|and 47∈π(K).SetP47∈Syl47(K).SinceG=KNG(P47),thenp||NG(P47)|.It is easy to check by Table 1 that there exists no such a primepsuch thatp||NG(P47)|,a contradiction.

Till now we have proved thatKis a-group whilep=157,271 or 331.Assume thatp∈{353,367,443,571,593,601,653,751,947,971}.Now,we have to discuss ten cases.IfKis aq-group for everyq∈π(G){2,3,5,p},it is easy to know that this is impossible by checking each choice ofpone by one.Since the methods used below are the same as in Subcase 1.3,we omit the detailed processes.ThereforeKis a{2,3,5}-group.Sinceit follows at once thatGis nonsolvable.This completes the proof of Case 1.

Case 2The quotient groupG/Kis an almost simple group.In other words,there exists a non-Abelian simple groupSsuch that

ProofLet:=G/KandS:=ThenS=B1×B2×B3×···×Bn,whereBj(j=1,2,···,n)are non-Abelian simple groups andAut(S).We assert thatn=1 andS=B1.

Suppose thatn≥2.We first show thatpdoes not divide the order ofS.If not,there exists a primersuch thatr∈ π(S)andr∈/π(K).Hence deg(p)≥4,a contradiction.Therefore,for eachj,one has thatBj∈Fp.On the other hand,by Lemma 3.7,we see thatp∈π(Aut(S)).Thuspdivides the order of Out(S).But Out(S)=Out(S1)×Out(S2)×···×Out(Sr),where the groupsSjare direct products of all isomorphicssuch thatS=S1×S2×···×Sr.Therefore for somei,pdivides the order of an outer automorphism group of a direct productSiofmisomorphic simple groupsBjfor some 1≤j≤n.SinceBj∈Fp,it follows that|Out(Bj)|is not divided bypby Lemma 3.7.Now,by Lemma 3.8,we obtain|Aut(Si)|=|Aut(Bj)|m·m!.Thereforem≥pand so 22pmust divide the order ofG.However,Exp(|Ap+5|,2)

Case 3In other words,Ap+5isOD-characterizable.

ProofBy Lemma 3.7 and Case 1,assume that·2α1·3α2·5α3,where 2≤α1≤|G|2=Exp(|Ap+5|,2)=u1,1≤α2≤|G|3=Exp(|Ap+5|,3)=u2,1≤α3≤|G|5=Exp(|Ap+5|,5)=u3.Letp1,p2,p3,···,psbe distinct consecutive prime numbers and 2=p1<3=p2<5=p3<···

IfSAp,thenHence,3·pa contradiction.

For the same reason,orAp+2.Therefore,Sis isomorphic to one of the simple groups:Ap+3,Ap+4andAp+5.

Letqbe an odd prime and 5

Table 2 E(q,p)

Ifp∈{131,173,167,233,257,263,271,331,353,367,373,433,443,503,541,557,563,571,653,587,607,677,733,941,947,977},by Table 2,Scan not be isomorphic to the simple groupAp+3orAp+4.Otherwise,there exists at least a primeqwith 5

Ifp∈{157,251,383,593,601,647,727,751,971},by Table 2 and Case 1,Kis a{2,3}-group.Inthis case,wherem=p+3 orp+4.By Case 2,we have thatAm≤G/K≤Aut(Am)But 5·p∈πe(G/K)πe(Sn),a contradiction.

Hence,By Case 2,one has thatAp+5Aut(Ap+5)Since|G|=|Ap+5|,IfG/KAp+5,then by comparing the orders we deduce thatAp+5,which completes the proof of Case 3 and also the proof of Theorem 2.1.

In 1989,Shi W.J.put forward the following conjecture.

Corollary 4.1(see[26])Let G be a group and M a finite simple group.Then GM if and only if(1)|G|=|M|and(2)πe(G)=πe(M).

The above conjecture was proved by joint works of many mathematicians and the last part of the proof was given by Mozurov V.D.etc.in[27].That is,the following theorem holds.

Theorem 4.1(see[27])Let G be a group and M a finite simple group.Then if and only if(1)|G|=|M|and(2)πe(G)=πe(M).

About the relation of Theorem 4.1 andOD-characterizable groups,we have the following facts:For two finite groupsGandM,ifπe(G)=πe(M),thenGandMmust have the same prime graph.Hence they have the same degree pattern.Therefore,we can get the following corollary by Theorem 2.1.

Corollary 4.2If G is a finite group such that(1)|G|=|Ap+5|and(2)πe(G)=πe(Ap+5),where5p∈π(1000!),

5 OD-Characterization of the Symmetric Group Sp+5

As we already mentioned,the symmetric groupsSpandSp+1,wherepis a prime,areOD-characterizable.Proposition 1.5 says that the symmetric groupsSnwithn≤100 andp,p+1 are 3-foldOD-characterizable.On the other hand,according to Proposition 1.6,S10is 8-foldOD-characterizable,andS10is the first symmetric group which is notOD-characterizable.Till now,we have not found a symmetric groupSn(np,p+1),exceptS10,which is not 3-foldOD-characterizable.Hence,it is an interesting and difficult topic to investigate how many-foldOD-characterization of symmetric groups are.Therefore,the first author of this article put forward the following conjecture.

Conjecture 5.1All the symmetric groupsSn(np,p+1),exceptS10,are 3-foldOD-characterizable.

In this section,we are going to give an affirmative answer to this conjecture for the symmetric groupSp+5.In other words,we will prove Theorem 2.2.

Proof of Theorem 2.2LetGbe a finite group satisfying|G|=|Sp+5|andD(G)=D(Sp+5),wherep+4 is a composite number,p+6 is a prime and 5p∈π(1000!).By[17],we only need to discuss the primespsuch thatp+4 is a composite number,p+6 is a prime and 100p+5}∩πe(G)=?,wherer,s∈π(G).By Lemma 3.4(2),deg(2)=|π(G)|? 1,so the prime graph ofGis connected.By the structure ofD(G),it is easy to check by the Magma software that Γ(G)= Γ(Sp+5).

LetKdenote the maximal normal solvable subgroup ofG.For the similar reason as the proof of Theorem 2.1,Kis a{2,3,5}-group andAut(Ap+5)Sp+5.HenceG/KAp+5orSp+5.IfG/Kthen by comparing the order we get thatIfG/KAp+5,then|K|=2 andK≤∩Z(G).ThereforeGis a central extension ofZ2byAp+5.IfGis a non-split extension ofZ2byAp+5,thenIfGis a split extension ofZ2byAp+5,then

We omit the details forSp+5because the arguments are quite similar to those forAp+5.We only mention that the non-isomorphic groupsZ2·Ap+5andZ2×Ap+5have the same order and the degree pattern asSp+5.HenceSp+5is 3-foldOD-characterizable,which completes the proof of Theorem 2.2.

AcknowledgementThe authors would like to express his deep gratitude to the referee for his or her invaluable comments and suggestions which helped to improve the paper.

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